# Proof: By Induction

We conduct a proof by induction.

### Base Case $n=0$

• We have to show that $X_0\not\in X_0.$
• By the axiom of foundation, the singleton $\{X_0\}$ contains an element that is disjoint from itself.
• This can only be the set $X_0$.
• Since $X_0\cap \{X_0\}=\emptyset$, we have $X_0\not\in X_0.$

### Induction step $n\rightarrow n+1$

• By the base case there is no such chain which fulfills $X_0\in X_1\in\ldots\in X_{n-1}\in X_n\in X_0.$
• If a chain $X_0\in X_1\in\ldots\in X_n\in X_{n+1}\in X_0$ existed, then there would be a set $X_{n+1}$ with $X_{n+1}\in X_0$ and $X_n\in X_{n+1}$.
• Therefore, we cound replace $X_n$ by $X_{n+1}$ and get the chain $X_0\in X_1\in\ldots\in X_{n-1}\in X_{n+1}\in X_0.$
• But this is a contradiction to the base case.

Github: ### References

#### Bibliography

1. Hoffmann, Dirk W.: "Grenzen der Mathematik - Eine Reise durch die Kerngebiete der mathematischen Logik", Spektrum Akademischer Verlag, 2011
2. Ebbinghaus, H.-D.: "Einführung in die Mengenlehre", BI Wisschenschaftsverlag, 1994, 3th Edition